Ethanol Production Matlab Help

Ethanol Production

Engineers in The food chemical industries use fermentation in many processes. The following equation describes Baker’s yeast fermentation.
a (C6 HI 206) + b (02) + c (NH3)
– C6 HI ON 03 + d (H 20) + e (C 02) + f (C 2 H 60)
The variables a, b, …• f represent the masses of the-products involved in the reaction.
In this formula C 6 HI 206 represents glucose, C 6 HI ON 03 represents yeast, and C 2 H 60

represents ethanol. This reaction produces ethanol, in addition to water and carbon dioxide. We want to determine the amount of ethanol 1 produced. The number of C, 0, N, and H atoms on the left must balance those on the right side of the equation. This gives four equations
6a = + e + 21
6a + 2b = 3 + d + 2e + 1
c=1
12a + 3c = 10+ 2d + 61

The fermentor is equipped with an oxygen sensor and a carbon dioxide sensor. These enable us to compute the respiratory quotient R:

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Thus the fifth equation is Rb – e = O.The yeast yield Y (grams of yeast produced per gram of glucose consumed) is related to a as follows.
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where 144 is the molecular weight of yeast and ISO is the molecular weight of glucose. By measuring the yeast yield Y we can compute a as follows: a = 144/ 180Y. This is the sixth equation.
Write a user-defined function that computes f, the amount of ethanol produced, with Rand Y as the function’s arguments. Test your function for two cases where Y is measured
to be 0.5: (a) R = 1.1 and (b) R = 1.05.
• Solution

First note that there are only four unknowns because the third equation directly gives c = I, and the sixth equation directly gives a :=; 144 / 180Y To write these equations in matrix form,let XI = b, X2 = d, X3 = e, and X4 = I. Then the equations can be written as

-X3 – 2X4 = 6 – 6(144/1S0Y)
2xI – X2 – 2X3 – X4 = 3 – 6(l44/180y)
-2X2 – 6X4 = 7 – 12(l44/180y)
RXI -X3 = 0

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The function tile is shown below.
function E = ethanol(R,Y)
% Computes ethanol produced from yeast reaction.
A = [0,0,-1,-2;2,-1,-2,-1; …
0,-2,0,-6;R,0,-1,0];
b = [6-6*(144./(180*Y));3-6*(144./(180*Y)); …
7-12* (144./ (180*Y)); 0] ;
x = A\b;
E = x(4);
The session is as follows:
»ethanol(1.1,0.5)
ans =
0.0654
»ethanol(1.05,0.5)
ans =
-0.0717

The negative value for E in the second case indicates that ethanol is being consumed rather
than produced,

Posted on July 30, 2015 in Linear Algebric Equations

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